Write a program to check if the given number is prime or not.
Prime Number: A number that is divisible by 1 and itself is a prime number.
Note: 1 is not a prime number.
Example:
Input: 13
Output: 13 is prime number
Explanation: If number only divisible by 1 and itself then number is prime.
Approach: Iterate till n-1 if the number is divisible by any number then the number is not prime, else the number is prime.
C
#include <stdio.h>
int main()
{
int n = 13;
if (n <= 1)
printf("Number is not prime ");
else
{
int flag = 0;
for (int i = 2; i < n; i++)
{
if (n % i == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
{
printf("Number is prime ");
}
else
{
printf("Number is not prime ");
}
}
return 0;
}
Java
public class CheckIsPrime{
public static void main(String[] args) {
int number=13;
if(number>1 && checkIsPrime(number))
System.out.println(number+" is prime number");
else
System.out.println(number+" is not prime number");
}
//Method to check prime number
private static boolean checkIsPrime(int number) {
for(int i=2;i<number;i++)
{
// If a number is divisible by any number from 2 to n-1
// then it is not a prime
if(number%i==0)
return false;
}
return true;
}
}
//Time Complexity:O(n)
//Space Complexity:O(1)
C++
#include <bits/stdc++.h>
using namespace std;
//Function to check for prime numbers
bool checkPrime(int n)
{
if(n==1)
return false;
for(int i=2;i<n;i++)
{
//If a number is divisible by
// any number from 2 to n-1
// then it is not a prime
if(n%i==0)
return false;
}
return true;
}
int main()
{
int n=13;
if(checkPrime(n))
cout<<n<<" is a prime\n";
else
cout<<n<<" is not a prime\n";
}
//Time Complexity :O(n)
//Space Complexity:O(1)